review: right hand rules
Post on 28-Mar-2022
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2P18-
Review:Right Hand Rules
1. Torque: Thumb = torque, fingers show rotation2. Feel: Thumb = I, Fingers = B, Palm = F3. Create: Thumb = I, Fingers (curl) = B4. Moment: Fingers (curl) = I, Thumb = Moment
4P18-
Magnetic Dipole MomentsAnµ IIA ≡≡ ˆ
Generate:
Feel:1) Torque to align with external field2) Forces as for bar magnets (seek field)
-DipoleU = ⋅µ B
6P18-
Put a Frog in a 16 T Magnet…
For details: http://www.hfml.sci.kun.nl/levitate.html
8P18-
Para/Ferromagnetism
Applied external field B0 tends to align the atomic magnetic moments (unpaired electrons)
9P18-
DiamagnetismEverything is slightly diamagnetic. Why? More later.
If no magnetic moments (unpaired electrons) then this effect dominates.
11P18-
Levitating a Diamagnet1) Create a strong field
(with a field gradient!)2) Looks like a dipole field3) Toss in a frog (diamagnet)4) Looks like a bar magnet
pointing opposite the field5) Seeks lower field (force up)
which balances gravity
SNN
S
Most importantly, its stable:Restoring force always towards the center
SN
12P18-
Using ∇B to Levitate
For details: http://www.hfml.ru.nl/levitation-movies.html
•Frog•Strawberry•Water Droplets•Tomatoes•Crickets
15P18-
Perfect Diamagnetism:“Magnetic Mirrors”
N
S
N
S
No matter what the angle, it floats -- STABILITY
16P18-
Using ∇B to Levitate
For details: http://www.hfml.sci.kun.nl/levitate.html
A Sumo Wrestler
20P18-
The Biot-Savart LawCurrent element of length ds carrying current Iproduces a magnetic field:
20 ˆ
4 rdI rsBd ×
=πµ
22P18-
Gauss’s Law – The Idea
The total “flux” of field lines penetrating any of these surfaces is the same and depends only on the amount of charge inside
23P18-
Ampere’s Law: The IdeaIn order to have a B field around a loop, there must be current punching through the loop
24P18-
Ampere’s Law: The Equation
The line integral is around any closed contour bounding an open surface S.
Ienc is current through S:
∫ =⋅ encId 0µsB
encS
I d= ⋅∫ J A
26P18-
Biot-Savart vs. Ampere
Biot-Savart Law
general current sourceex: finite wire
wire loop
Ampere’s law
symmetriccurrent source
ex: infinite wireinfinite current sheet
02
ˆ4I d
rµπ
×= ∫
s rB
∫ =⋅ encId 0µsB
27P18-
Applying Ampere’s Law1. Identify regions in which to calculate B field
Get B direction by right hand rule2. Choose Amperian Loops S: Symmetry
B is 0 or constant on the loop!3. Calculate 4. Calculate current enclosed by loop S5. Apply Ampere’s Law to solve for B
∫ ⋅ sB d
∫ =⋅ encId 0µsB
28P18-
Always True, Occasionally Useful
Like Gauss’s Law, Ampere’s Law is always true
However, it is only useful for calculation in certain specific situations, involving highly symmetric currents. Here are examples…
29P18-
Example: Infinite Wire
I A cylindrical conductor has radius R and a uniform current density with total current I
Find B everywhere
Two regions:(1) outside wire (r ≥ R)(2) inside wire (r < R)
31P18-
Example: Wire of Radius RRegion 1: Outside wire (r ≥ R)
d⋅∫ B s
ckwisecounterclo2
0
rI
πµ
=B
B d= ∫ s ( )2B rπ=
0 encIµ= 0Iµ=
Cylindrical symmetry Amperian CircleB-field counterclockwise
32P18-
Example: Wire of Radius RRegion 2: Inside wire (r < R)
2
0 2
rIR
πµπ⎛ ⎞
= ⎜ ⎟⎝ ⎠
ckwisecounterclo2 2
0
RIr
πµ
=B
Could also say: ( )222 ; rRIJAI
RI
AIJ encenc π
ππ====
d⋅∫ B s B d= ∫ s ( )2B rπ=
0 encIµ=
34P18-
Group Problem: Non-Uniform Cylindrical Wire
I A cylindrical conductor has radius R and a non-uniform current density with total current:
Find B everywhere
0RJr
=J
35P18-
Applying Ampere’s LawIn Choosing Amperian Loop:• Study & Follow Symmetry• Determine Field Directions First• Think About Where Field is Zero• Loop Must
• Be Parallel to (Constant) Desired Field• Be Perpendicular to Unknown Fields• Or Be Located in Zero Field
41P18-
Magnetic Field of Solenoid
loosely wound tightly wound
For ideal solenoid, B is uniform inside & zero outside
42P18-
Magnetic Field of Ideal Solenoid
d d d d d⋅ ⋅ + ⋅ + ⋅ + ⋅∫ ∫ ∫ ∫ ∫1 2 3 4
B s = B s B s B s B s
Using Ampere’s law: Think!
along sides 2 and 4
0 along side 3
d⎧ ⊥⎪⎨
=⎪⎩
B sB
n: turn densityencI nlI=
0d Bl nlIµ⋅ = =∫ B s
00
nlIB nIl
µ µ= =/ : # turns/unit lengthn N L=
0 0 0Bl= + + +
44P18-
Group Problem: Current Sheet
A sheet of current (infinite in the y & z directions, of thickness 2d in the x direction) carries a uniform current density:
Find B everywhere
ˆs J=J k
y
45P18-
Ampere’s Law:Infinite Current Sheet
I
Amperian Loops:B is Constant & Parallel OR Perpendicular OR Zero
I Penetrates
B
B
46P18-
Solenoid is Two Current SheetsField outside current sheet should be half of solenoid, with the substitution:
2nI dJ=
This is current per unit length (equivalent of λ, but we don’t have a symbol for it)
47P18-
=2 Current Sheets
Ampere’s Law: .∫ =⋅ encId 0µsB
IB
B
X XX
X
X
XX
X
XXX
X
X
XX
X
XXXXXXXXXXXX
B
LongCircular
Symmetry(Infinite) Current Sheet
Solenoid
Torus
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